$A=\left[\begin{array}{rr}2 & 6 & 9 \\3 & 4 & 1 \\7 &7 &8\end{array}\right]$ $A_{3,3}=$
Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{3,3}$ $A_{{3},{3}}$ is located on row ${3}$ of $A$ : $\left[\begin{array}{rr}2 & 6 & 9 \\3 & 4 & 1 \\ {7} & {7} & {8} \end{array}\right]$ $A_{{3},{3}}$ is also located on column ${3}$ of $A$. $\left[\begin{array}{rr}2 & 6 & {9} \\3 & 4 & {1} \\ {7} & {7} & {\text8} \end{array}\right]$ Therefore, $A_{{3},{3}}={8}$. Summary $A_{3,3}=8$